$$ \text { Match List I with List II. } $$ $$ \text { List-I } $$ $$ \text { List-II }

A. $E=h v$ is energy of photon B. Diffraction and interference confirm wave nature of light C. $\lambda=\frac{h}{p}$ is de Broglie wavelength of particle. D. Compton effect confirms particle nature of light.

Dual Nature of Radiation and Matter Questions — NEET Physics

Q1

<p>

\text{ Match List I with List II. }

</p> <table class=tg><thead> <tr> <th class=tg-c3ow colspan=2>

<br>\text{ List-I }<br>

<br>\text{ List-II }<br>

E=h v

</td> <td class=tg-c3ow>I.</td> <td class=tg-c3ow>de Broglie wavelength</td> </tr> <tr> <td class=tg-c3ow>B.</td> <td class=tg-c3ow>Diffraction and Interference</td> <td class=tg-c3ow>II.</td> <td class=tg-c3ow>Particle nature of light</td> </tr> <tr> <td class=tg-c3ow>C.</td> <td class=tg-c3ow>

\lambda=h / p

</td> <td class=tg-c3ow>III.</td> <td class=tg-c3ow>Wave nature of light</td> </tr> <tr> <td class=tg-c3ow>D.</td> <td class=tg-c3ow>Compton effect</td> <td class=tg-c3ow>IV.</td> <td class=tg-c3ow>Energy of photon</td> </tr> </tbody> </table>

A A-IV, B-I, C-II, D-III

B A-IV, B-III, C-II, D-I

C A-I, B-IV, C-III, D-II

D A-IV, B-III, C-I, D-II

Correct Answer

Option D

Solution

A. $E=h v$ is energy of photon B.

Diffraction and interference confirm wave nature of light C. $\lambda=\dfrac{h}{p}$ is de Broglie wavelength of particle.

D.

Compton effect confirms particle nature of light.

Q2

For a metal of work function 6.6 eV , which of the following wavelengths of incident radiation does not give rise to the photoelectric effect? (Take Planck's constant as

6.6 \times 10^{-34} \mathrm{~J} \mathrm{~s}

)

A 100 nm

B 150 nm

C 200 nm

D 50 nm

Correct Answer

Option C

Solution

For incident radiation having wavelength ( $\lambda$ ), photoelectric effect doesn't occur when $\frac{h c}{\lambda}

\begin{aligned} & \Rightarrow \lambda>\frac{h c}{W_0} \\ & \Rightarrow \lambda>\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{6.6 \times 1.6 \times 10^{-19}} \\ & \Rightarrow \lambda>\frac{3 \times 10^{-7}}{1.6} \\ & \Rightarrow \lambda>\frac{300}{1.6} \mathrm{~nm} \\ & \Rightarrow \lambda=187.5 \mathrm{~nm} \end{aligned}

∴ Option (3) 200 nm is correct.

Q3

De-Broglie wavelength of an electron orbiting in the

n=2

state of hydrogen atom is close to (Given Bohr radius

=0.052 \mathrm{~nm}

)

A 1.67 nm

B 2.67 nm

C 0.067 nm

D 0.67 nm

Correct Answer

Option D

Solution

The De-Broglie wavelength of an electron in the $n=2$ state of a hydrogen atom can be calculated as follows: Firstly, calculate the radius $r$ for the $n=2$ state using the Bohr radius formula: $r = 0.052 n^2$ For $n=2$ , the radius is: $\begin{aligned} r &= 0.052 \times 4 \\ &= 0.208 \, \text{nm} \end{aligned}$ We use the relationship $M v r = \dfrac{n h}{2 \pi}$ , leading to the expression for the De-Broglie wavelength $\lambda$ : $\lambda = \dfrac{h}{M v} = \pi r$ Substituting the calculated $r$ value: $\begin{aligned} \lambda &= \pi \times 0.208 \, \text{nm} \\ &= 3.14 \times 0.208 \, \text{nm} \\ &= 0.65317 \, \text{nm} \\ &\approx 0.67 \, \text{nm} \end{aligned}$ Thus, the De-Broglie wavelength of the electron in the $n=2$ state is approximately $0.67$ nm.

Q4

Which of the following options represent the variation of photoelectric current with property of light shown on the x-axis?

A A and D

B B and D

C A only

D A and C

Correct Answer

Option C

Solution

Photoelectric current is directly proportional to intensity of light.

Q5

A photon and an electron (mass

m

) have the same energy

E

. The ratio (

\lambda_{\text{photon }} / \lambda_{\text{electron }}

) of their de Broglie wavelengths is: (

c

is the speed of light)

A

c \sqrt{\dfrac{2 m}{E}}

B

\dfrac{1}{c} \sqrt{E / 2 m}

C

\sqrt{E / 2 m}

D

c \sqrt{2 m E}

Correct Answer

Option A

Solution

To find the ratio of the de Broglie wavelengths of a photon and an electron when they both have the same energy $E$ , follow these steps: Photon Wavelength: For a photon, the energy is given by: $E = \dfrac{h c}{\lambda_{\mathrm{Ph}}}$ Where $h$ is Planck's constant, $c$ is the speed of light, and $\lambda_{\mathrm{Ph}}$ is the wavelength of the photon.

Rearranging this equation gives: $\lambda_{\mathrm{Ph}} = \dfrac{h c}{E}$ Electron Wavelength: For an electron, the energy related to its momentum is: $E = \dfrac{p^2}{2m}$ Where $m$ is the mass of the electron and $p$ is its momentum.

Using the de Broglie wavelength expression $p = \dfrac{h}{\lambda_e}$ , we can write: $E = \left( \dfrac{h}{\lambda_e} \right)^2 \times \dfrac{1}{2m}$ Solving for the electron's wavelength $\lambda_e$ : $\lambda_e = \dfrac{h}{\sqrt{2mE}}$ Ratio of Wavelengths: Now, calculate the ratio of the wavelengths $\dfrac{\lambda_{\mathrm{Ph}}}{\lambda_e}$ as follows: $\dfrac{\lambda_{\mathrm{Ph}}}{\lambda_e} = \dfrac{\dfrac{h c}{E}}{\dfrac{h}{\sqrt{2mE}}} = c \sqrt{\dfrac{2m}{E}}$ This ratio equation defines the relation between the photon and electron wavelengths when both have the same energy $E$ .

Q6

An electron and an alpha particle are accelerated by the same potential difference. Let

\lambda_e

and

\lambda_\alpha

denote the de-Broglie wavelengths of the electron and the alpha particle, respectively, then:

A

\lambda_e>\lambda_\alpha

B

\lambda_e=4 \lambda_\alpha

C

\lambda_e=\lambda_\alpha

D

\lambda_e<\lambda_\alpha

Correct Answer

Option A

Solution

de-Broglie wavelength is given by

\lambda=\frac{h}{p}=\frac{h}{\sqrt{2 m q V}}

For same potential difference

\begin{gathered} \lambda \propto \frac{1}{\sqrt{m q}} \\ \frac{\lambda_\alpha}{\lambda_e}=\sqrt{\frac{m_e q_e}{m_\alpha q_\alpha}} \\ \because m_\alpha \gg m_e \\ \lambda_e>\lambda_\alpha \end{gathered}

Q7

If

\phi

is the work function of photosensitive material in

\mathrm{eV}

and light of wavelength of numerical value

\lambda=\dfrac{h c}{e}

metre, is incident on it with energy above its threshold value at an instant then the maximum kinetic energy of the photo-electron ejected by it at that instant (Take

h

-Plank's constant,

c

-velocity of light in free space) is (in SI units):

A

e+2 \phi

B

2 e-\phi

C

e-\phi

D

e+\phi

Correct Answer

Option C

Solution

The energy of the incident light is given by:

E = h \nu = \frac{hc}{\lambda}

where

h

is Planck's constant,

c

is the speed of light, and $\lambda$ is the wavelength of the light. Substituting the given value of $\lambda$ , we get:

E = \frac{hc}{hc/e} = e

The maximum kinetic energy of the photoelectron is given by:

KE_{max} = E - \phi

where $\phi$ is the work function. Substituting the values, we get:

KE_{max} = e - \phi

Therefore, the correct answer is Option C .

Q8

The de Broglie wavelength associated with an electron, accelerated by a potential difference of 81 V is given by:

A 13.6 nm

B 136 nm

C 1.36 nm

D 0.136 nm

Correct Answer

Option D

Solution

{\lambda _e} = {{12.27} \over {\sqrt V }}\mathop A\limits^o = {{12.27} \over {\sqrt {81} }}\mathop A\limits^o = {{12.27} \over 9}\mathop A\limits^o

= 1.36\mathop A\limits^o

( $\because$

1\mathop A\limits^o = {1 \over {10}}

nm)

= 0.136

nm

Q9

The maximum kinetic energy of the emitted photoelectrons in photoelectric effect is independent of:

A work function of material

B intensity of incident radiation

C frequency of incident radiation

D wavelength of incident radiation

Correct Answer

Option B

Solution

Maximum kinetic energy of emitted electron is independent of intensity of radiation.

Q10

The work functions of Caesium

(\mathrm{Cs})

, potassium

(\mathrm{K})

and Sodium (Na) are

2.14 ~\mathrm{eV}, 2.30 ~\mathrm{eV}

and

2.75 ~\mathrm{eV}

respectively. If incident electromagnetic radiation has an incident energy of

2.20 ~\mathrm{eV}

, which of these photosensitive surfaces may emit photoelectrons?

A Both Na and K

B K only

C Na only

D Cs only

Correct Answer

Option D

Solution

Given energy of photon

E=2.20 ~\mathrm{eV}

Work function of

\mathrm{Cs} ~\phi_{0}=2.14 ~\mathrm{eV}, \mathrm{K} ~\phi_{0}=2.30 ~\mathrm{eV}, \mathrm{Na} ~\phi_{0}=2.75 ~\mathrm{eV}

We know that

e^{-}

emitts when

h v>\phi_{0}

here it is clear that energy of photon is more than the work function of

\mathrm{Cs}

[Caesium] only so Ans. only (Cs).